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Q: Who created the B 2 Stealth?
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How do the squares of sums and differences relate?

(a + b) = a^2 + 2ab + b^2 (a - b)^2 = a^2 - 2ab + b^2 or you can work like this: [a + (-b)]^2 = a^2 + 2a(-b) + (-b)^2 (a - b)^2 = a^2 - 2ab + b^2


What are the different special product formulas?

1. Square of a binomial (a+b)^2 = a^2 + 2ab + b^2 carry the signs as you solve 2. Square of a Trinomial (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc carry the sings as you solve 3. Cube of a Binomial (a+b)^3 = a^3 + 3(a^2)b + 3a(b^2) + b^3 4. Product of sum and difference (a+b)(a-b) = a^2 - b^2 5. Product of a binomial and a special multinomial (a+b)(a^2 - ab + b^2) = a^3-b^3 (a-b)(a^2 + ab + b^2) = a^3-b^3


What is proof of Heron's Formula?

This is a proof that uses the cosine rule and Pythagoras' theorem. As on any triangle with c being the opposite side of θ and a and b are the other sides: c^2=a^2+b^2-2abcosθ We can rearrange this for θ: θ=arccos[(a^2+b^2-c^2)/(2ab)] On a right-angle triangle cosθ=a/h. We can therefore construct a right-angle triangle with θ being one of the angles, the adjacent side being a^2+b^2-c^2 and the hypotenuse being 2ab. As the formula for the area of a triangle is also absinθ/2, when a and b being two sides and θ the angle between them, the opposite side of θ on the right-angle triangle we have constructed is 4A, with A being the area of the original triangle, as it is 2absinθ. Therefore, according to Pythagoras' theorem: (2ab)^2=(a^2+b^2-c^2)^2+(4A)^2 4a^2*b^2=(a^2+b^2-c^2)^2+16A^2 16A^2=4a^2*b^2-(a^2+b^2-c^2)^2 This is where it will start to get messy: 16A^2=4a^2*b^2-(a^2+b^2-c^2)(a^2+b^2-c^2) =4a^2*b^2-(a^4+a^2*b^2-a^2*c^2+a^2*b^2+b^4-b^2*c^2- a^2*c^2-b^2*c^2+c^4) =4a^2*b^2-(a^4+2a^2*b^2-2a^2*c^2+b^4-2b^2*c^2+c^4) =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.1) We will now see: (a+b+c)(-a+b+c)(a-b+c)(a+b-c) =(-a^2+ab+ac-ab+b^2+bc-ac+bc+c^2)(a^2+ab-ac-ab-b^2+bc+ac+bc-c^2) =(-a^2+b^2+2bc+c^2)(a^2-b^2+2bc-c^2) =-a^4+a^2*b^2-2a^2*bc+a^2*c^2+a^2*b^2-b^4+2b^3*c-b^2*c^2+2a^2*bc-2b^3*c+(2bc)^2-2bc^3+a^2*c^2-b^2*c^2+2bc^3-c^4 =-a^4+2a^2*b^2+2a^2*c^2-b^4+(2bc)^2-c^4-2b^2*c^2 =-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.2) And now that we know that Eq.1=Eq.2, we can make Eq.1=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) Therefore: 16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16 =[(a+b+c)/2][(-a+b+c)/2][(a-b+c)/2][(a+b-c)/2] And so if we let s=(a+b+c)/2 A^2=s(s-a)(s-b)(s-c)


How do you prove 2 equals 1?

It can if you divide by zero. 1. Let a and b be equal non-zero quantities a = b 2. Multiply both sides by a a^2 = ab 3. Subtract b^2 a^2 - b^2 = ab - b^2 4. Factor both sides (a - b)(a + b) = b(a - b) 5. Divide out (a - b) a + b = b 6. Since a = b ... b + b = b 7. Combine like terms on the left 2b = b 8. Divide by the non-zero b 2 = 1


A and b are numbers where a equals b plus 2 the sum of a and v is equal to the product of a and b show that a and b are not integers?

Basically, you have the following linear equation system: a = b + 2 a + b = a * b (I assume "v" is a typo for "b"). Substituting (b+2) for a in the second equation, we get: 2b+2 = b(b+2) = b^2+2b Subtracting (2b) from both sides, we get: 2 = b^2 and we can conclude that: b = sqrt(2) which is not an integer (in fact, it is not even rational). And it is pretty easy to prove that sqrt(2) is not an integer, if needed.

Related questions

Who created the B-2 Stealth?

A design team from Northrop Grumman and Boeing.


What division created the b2?

Type your answer here... The B-2 Stealth bomber was created proudly in the U.S Air Force branch of the military


What company put the stealth in the stealth bomber?

Northrop Grumman built the B-2.


What company that put the stealth in the stealth bomber?

Northrop Grumman built the B-2.


What is the best stealth bomber?

At the moment, the B-2 Spirit is the only stealth bomber in service.


Who invented the Stealth Bomber B-2?

See website: B-2 Bomber


Where was the B-2 Stealth Bomber built at?

Area 51


What year was the b-12 stealth bomber invented?

See website: B-2 bomber


Where and how do you purchase a authentic B-2 Stealth Bomber?

Simple answer, you can't.


What is the real name of the stealth bomber in Modern Warfare 2?

B-2 Spirit


Who was the first person to fly the stealth bomber b-2?

tom hickens


When was Wichita Stealth created?

Wichita Stealth was created in 1999.